Stern-Brocot sequence
For this task, the Stern-Brocot sequence is to be generated by an algorithm similar to that employed in generating the Fibonacci sequence.
1. The first and second members of the sequence are both 1
1, 1
2. Start by considering the second member of the sequence
3. Sum the considered member of the sequence and its precedent, (1 + 1) = 2, and append it to the end of the sequence.
1, 1, 2
4. Append the considered member of the sequence to the end of the sequence.
1, 1, 2, 1
5. Consider the next member of the series, (the third member i.e. 2)
6. GOTO 3
Expanding another loop we get:
7. Sum the considered member of the sequence and its precedent, (2 + 1) = 3, and append it to the end of the sequence.
1, 1, 2, 1, 3
8. Append the considered member of the sequence to the end of the sequence.
1, 1, 2, 1, 3, 2
9. Consider the next member of the series, (the fourth member i.e. 1)
- The task is to
- Create a function/method/subroutine/procedure/... to generate the Stern-Brocot sequence of integers using the method outlined above.
- Show the first fifteen members of the sequence. (This should be: 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4)
- Show the (1-based) index of where the numbers 1-to-10 first appears in the sequence.
- Show the (1-based) index of where the number 100 first appears in the sequence.
- Check that the greatest common divisor of all the two consecutive members of the series up to the 1000th member, is always one.
Show your output on the page.
- Ref
- Infinite Fractions - Numberphile (Video).
- Trees, Teeth, and Time: The mathematics of clock making.
- A002487 The On-Line Encyclopedia of Integer Sequences.
- Related Tasks
C
Recursive function. <lang c>#include <stdio.h>
typedef unsigned int uint;
/* the sequence, 0-th member is 0 */ uint f(uint n) { return n < 2 ? n : (n&1) ? f(n/2) + f(n/2 + 1) : f(n/2); }
uint gcd(uint a, uint b) { return a ? a < b ? gcd(b%a, a) : gcd(a%b, b) : b; }
void find(uint from, uint to) { do { uint n; for (n = 1; f(n) != from ; n++); printf("%3u at Stern #%u.\n", from, n); } while (++from <= to); }
int main(void) { uint n; for (n = 1; n < 16; n++) printf("%u ", f(n)); puts("are the first fifteen.");
find(1, 10); find(100, 0);
for (n = 1; n < 1000 && gcd(f(n), f(n+1)) == 1; n++); printf(n == 1000 ? "All GCDs are 1.\n" : "GCD of #%d and #%d is not 1", n, n+1);
return 0; }</lang>
- Output:
1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 are the first fifteen. 1 at Stern #1. 2 at Stern #3. 3 at Stern #5. 4 at Stern #9. 5 at Stern #11. 6 at Stern #33. 7 at Stern #19. 8 at Stern #21. 9 at Stern #35. 10 at Stern #39. 100 at Stern #1179. All GCDs are 1.
The above f() can be replaced by the following, which is much faster but probably less obvious as to how it arrives from the recurrence relations.
<lang c>uint f(uint n)
{
uint a = 1, b = 0;
while (n) {
if (n&1) b += a;
else a += b;
n >>= 1;
}
return b;
}</lang>
D
<lang d>import std.stdio, std.numeric, std.range, std.algorithm;
/// Generates members of the stern-brocot series, in order, /// returning them when the predicate becomes false. uint[] sternBrocot(bool delegate(in uint[]) pure nothrow @safe @nogc pred=seq => seq.length < 20) pure nothrow @safe {
typeof(return) sb = [1, 1];
size_t i = 0;
while (pred(sb)) {
sb ~= [sb[i .. i + 2].sum, sb[i + 1]];
i++;
}
return sb;
}
void main() {
enum nFirst = 15;
writefln("The first %d values:\n%s\n", nFirst,
sternBrocot(seq => seq.length < nFirst).take(nFirst));
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only))
writefln("1-based index of the first occurrence of %3d in the series: %d",
nOccur, sternBrocot(seq => nOccur != seq[$ - 2]).length - 1);
enum nGcd = 1_000;
auto s = sternBrocot(seq => seq.length < nGcd).take(nGcd);
assert(zip(s, s.dropOne).all!(ss => ss[].gcd == 1),
"A fraction from adjacent terms is reducible.");
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
This uses a queue from the Queue/usage Task: <lang d>import std.stdio, std.algorithm, std.range, std.numeric, queue_usage2;
struct SternBrocot {
private auto sb = GrowableCircularQueue!uint(1, 1);
enum empty = false;
@property uint front() pure nothrow @safe @nogc {
return sb.front;
}
uint popFront() pure nothrow @safe {
sb.push(sb.front + sb[1]);
sb.push(sb[1]);
return sb.pop;
}
}
void main() {
SternBrocot().drop(50_000_000).front.writeln;
}</lang>
- Output:
7004
Direct Version:
<lang d>void main() {
import std.stdio, std.numeric, std.range, std.algorithm, std.bigint, std.conv;
/// Stern-Brocot sequence, 0-th member is 0.
T sternBrocot(T)(T n) pure nothrow /*safe*/ {
T a = 1, b = 0;
while (n) {
if (n & 1) b += a;
else a += b;
n >>= 1;
}
return b;
}
alias sb = sternBrocot!uint;
enum nFirst = 15;
writefln("The first %d values:\n%s\n", nFirst, iota(1, nFirst + 1).map!sb);
foreach (immutable nOccur; iota(1, 10 + 1).chain(100.only))
writefln("1-based index of the first occurrence of %3d in the series: %d",
nOccur, sequence!q{n}.until!(n => sb(n) == nOccur).walkLength);
auto s = iota(1, 1_001).map!sb;
assert(s.zip(s.dropOne).all!(ss => ss[].gcd == 1),
"A fraction from adjacent terms is reducible.");
sternBrocot(10.BigInt ^^ 20_000).text.length.writeln;
}</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179 7984
Haskell
<lang haskell>import Data.List
sb = 1:1: f (tail sb) sb where f (a:aa) (b:bb) = a+b : a : f aa bb
main = do print $ take 15 sb print [(i,1 + (\(Just i)->i) (elemIndex i sb)) | i <- [1..10]++[100]] print $ all (\(a,b)->1 == gcd a b) $ take 1000 $ zip sb (tail sb)</lang>
- Output:
[1,1,2,1,3,2,3,1,4,3,5,2,5,3,4] [(1,1),(2,3),(3,5),(4,9),(5,11),(6,33),(7,19),(8,21),(9,35),(10,39),(100,1179)] True
J
We have two different kinds of list specifications here (length of the sequence, and the presence of certain values in the sequence). Also the underlying list generation mechanism is somewhat arbitrary. So let's generate the sequence iteratively and provide a truth valued function of the intermediate sequences to determine when we have generated one which is adequately long:
<lang J>sternbrocot=:1 :0
ind=. 0
seq=. 1 1
while. -. u seq do.
ind=. ind+1
seq=. seq, +/\. seq {~ _1 0 +ind
end.
)</lang>
First fifteen members of the sequence:
<lang J> 15{.(15<:#) sternbrocot 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4</lang>
One based indices of where numbers 1-10 first appear in the sequence:
<lang J> 1+(10 e. ]) sternbrocot i.1+i.10 1 3 5 9 11 33 19 21 35 39</lang>
One based index of where the number 100 first appears:
<lang J> 1+(100 e. ]) sternbrocot i. 100 1179</lang>
List of distinct greatest common divisors of adjacent number pairs from a sternbrocot sequence which includes the first 1000 elements:
<lang J> ~.2 +./\ (1000<:#) sternbrocot 1</lang>
Java
This example generates the first 1200 members of the sequence since that is enough to cover all of the tests in the description. It borrows the gcd method from BigInteger rather than using its own.
<lang java5>import java.math.BigInteger;
import java.util.LinkedList;
public class SternBrocot { static LinkedList<Integer> sequence = new LinkedList<Integer>()Template:Add(1); add(1);;
private static void genSeq(int n){ for(int conIdx = 1; sequence.size() < n; conIdx++){ int consider = sequence.get(conIdx); int pre = sequence.get(conIdx - 1); sequence.add(consider + pre); sequence.add(consider); }
}
public static void main(String[] args){ genSeq(1200); System.out.println("The first 15 elements are: " + sequence.subList(0, 15)); for(int i = 1; i <= 10; i++){ System.out.println("First occurrence of " + i + " is at " + (sequence.indexOf(i) + 1)); }
System.out.println("First occurrence of 100 is at " + (sequence.indexOf(100) + 1));
boolean failure = false; for(int i = 0; i < 999; i++){ failure |= !BigInteger.valueOf(sequence.get(i)).gcd(BigInteger.valueOf(sequence.get(i + 1))).equals(BigInteger.ONE); } System.out.println("All GCDs are" + (failure ? " not" : "") + " 1"); } }</lang>
- Output:
The first 15 elements are: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] First occurrence of 1 is at 1 First occurrence of 2 is at 3 First occurrence of 3 is at 5 First occurrence of 4 is at 9 First occurrence of 5 is at 11 First occurrence of 6 is at 33 First occurrence of 7 is at 19 First occurrence of 8 is at 21 First occurrence of 9 is at 35 First occurrence of 10 is at 39 First occurrence of 100 is at 1179 All GCDs are 1
Python
Python: procedural
<lang python>def stern_brocot(predicate=lambda series: len(series) < 20):
"""\ Generates members of the stern-brocot series, in order, returning them when the predicate becomes false
>>> print('The first 10 values:',
stern_brocot(lambda series: len(series) < 10)[:10])
The first 10 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3]
>>>
"""
sb, i = [1, 1], 0
while predicate(sb):
sb += [sum(sb[i:i + 2]), sb[i + 1]]
i += 1
return sb
if __name__ == '__main__':
from fractions import gcd
n_first = 15
print('The first %i values:\n ' % n_first,
stern_brocot(lambda series: len(series) < n_first)[:n_first])
print()
n_max = 10
for n_occur in list(range(1, n_max + 1)) + [100]:
print('1-based index of the first occurrence of %3i in the series:' % n_occur,
stern_brocot(lambda series: n_occur not in series).index(n_occur) + 1)
# The following would be much faster. Note that new values always occur at odd indices
# len(stern_brocot(lambda series: n_occur != series[-2])) - 1)
print()
n_gcd = 1000
s = stern_brocot(lambda series: len(series) < n_gcd)[:n_gcd]
assert all(gcd(prev, this) == 1
for prev, this in zip(s, s[1:])), 'A fraction from adjacent terms is reducible'</lang>
- Output:
The first 15 values: [1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4] 1-based index of the first occurrence of 1 in the series: 1 1-based index of the first occurrence of 2 in the series: 3 1-based index of the first occurrence of 3 in the series: 5 1-based index of the first occurrence of 4 in the series: 9 1-based index of the first occurrence of 5 in the series: 11 1-based index of the first occurrence of 6 in the series: 33 1-based index of the first occurrence of 7 in the series: 19 1-based index of the first occurrence of 8 in the series: 21 1-based index of the first occurrence of 9 in the series: 35 1-based index of the first occurrence of 10 in the series: 39 1-based index of the first occurrence of 100 in the series: 1179
Python: More functional
An iterator is used to produce successive members of the sequence. (its sb variable stores less compared to the procedural version aboveby popping the last element every time around the while loop.
In checking the gcd's, two iterators are tee'd off from the one stream with the second advanced by one value with its call to next().
See the talk page for how a deque was selected over the use of a straightforward list' <lang python>>>> from itertools import takewhile, tee, islice >>> from collections import deque >>> from fractions import gcd >>> >>> def stern_brocot():
sb = deque([1, 1])
while True:
sb += [sb[0] + sb[1], sb[1]]
yield sb.popleft()
>>> [s for _, s in zip(range(15), stern_brocot())]
[1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 5, 3, 4]
>>> [1 + sum(1 for i in takewhile(lambda x: x != occur, stern_brocot()))
for occur in (list(range(1, 11)) + [100])]
[1, 3, 5, 9, 11, 33, 19, 21, 35, 39, 1179] >>> prev, this = tee(stern_brocot(), 2) >>> next(this) 1 >>> all(gcd(p, t) == 1 for p, t in islice(zip(prev, this), 1000)) True >>> </lang>
REXX
This REXX program could've been made simplier by including a GCD function that ignores zeroes and negative numbers, and also only processes two numbers. <lang rexx>/*REXX pgm gens/shows Stern-Brocot sequence, finds 1-based indices, GCDs*/ parse arg N idx fix chk . /*get optional argument from C.L.*/ if N== | N==',' then N= 15 /*use the default for N ? */ if idx== | idx==',' then idx= 10 /* " " " " idx ? */ if fix== | fix==',' then fix= 100 /* " " " " fix ? */ if chk== | chk==',' then chk=1000 /* " " " " chk ? */
/*═══════════════════════════════*/
say center('the first' N 'numbers in the Stern-Brocot sequence', 70,'═') a=Stern_Brocot(N) /*invoke function to generate seq*/ say a /*display sequence to terminal. */
/*═══════════════════════════════*/
say center('the 1-based index for the first' idx "integers", 70, '═') a=Stern_Brocot(-idx) /*invoke function to generate seq*/
do i=1 for idx
say 'for ' right(i,length(idx))", the index is: " wordpos(i,a)
end /*i*/
/*═══════════════════════════════*/
say center('the 1-based index for' fix, 70, '═') a=Stern_Brocot(-fix) /*invoke function to generate seq*/ say 'for ' fix", the index is: " wordpos(fix,a)
/*═══════════════════════════════*/
say center('checking if all two consecutive members have a GCD=1', 70,'═') a=Stern_Brocot(chk) /*invoke function to generate seq*/
do c=1 for chk-1; if gcd(subword(a,c,2))==1 then iterate
say 'GCD check failed at member' c"."; exit 13
end /*c*/
say '───── All ' chk " two consecutive members have a GCD of unity." exit /*stick a fork in it, we're done.*/ /*──────────────────────────────────GCD subroutine──────────────────────*/ gcd: procedure; $=; do i=1 for arg(); $=$ arg(i); end /*arg list*/ parse var $ x z .; if x=0 then x=z /*handle special 0 case.*/ x=abs(x)
do j=2 to words($); y=abs(word($,j)); if y=0 then iterate
do until _==0; _=x//y; x=y; y=_; end /*◄──heavy lifting*/
end /*j*/
return x /*──────────────────────────────────STERN_BROCOT subroutine.────────────*/ Stern_Brocot: parse arg h 1 f; if h<0 then h=1e9; else f=0; f=abs(f) $=1 1
do k=2 until words($)>=h; _=word($,k); $=$ (_+word($,k-1)) _
if f==0 then iterate; if wordpos(f,$)\==0 then leave
end /*until*/
if f==0 then return subword($,1,h)
return $</lang>
output when using the default inputs:
══════════the first 15 numbers in the Sterm-Brocot sequence═══════════ 1 1 2 1 3 2 3 1 4 3 5 2 5 3 4 ═════════════the 1-based index for the first 10 integers══════════════ for 1, the index is: 1 for 2, the index is: 3 for 3, the index is: 5 for 4, the index is: 9 for 5, the index is: 11 for 6, the index is: 33 for 7, the index is: 19 for 8, the index is: 21 for 9, the index is: 35 for 10, the index is: 39 ══════════════════════the 1-based index for 100═══════════════════════ for 100, the index is: 1179 ═════════checking if all two consecutive members have a GCD=1═════════ ───── All 1000 two consecutive members have a GCD of unity.